please help to find the answer! Physics. Thank you?

Question

The energy required for a 440 kg mass to climb from the surface of the Earth to a point 6400 km 'above' the surface--i.e., at two Earth radii from the center--can be approximated by first averaging the field strength of the Earth at this point and at its surface.
•Given a surface field strength of 9.8 m/s^2, what is the field strength at two Earth radii from its center? What is the average of these two strengths?

•Using this average to approximate the actual average field strength determine the work necessary to climb from the surface to a point two Earth radii from its center.

Answer 1

jp needed an r², not r in his first equation. Using this correct variation, the field strength at 2r is 1/2² of the strength at the surface. The average is therefore (1+1/4)/2 = .625g

So the average force would be 440 kg*(.625g) = 2695 N and the work needed would be this times the height of 6.4E6 m:

= 1.7248E10 Joules

Note: an arithmetic average of a 2nd power function is going to be a VERY rough estimate...... Also, we didn't even take into consideration the rotation of the earth; a BIG factor.

Answer 2

10

Answer 3

((G)(M1)(M2))/(R) = gravitational acceleration equation
well just say for the hell of it ((G)(M1)(M2)) = 9.8
(R = 1 at the surface)
so:
Field strength at 2 radii= 50%
((9.8/R)+(9.8/(2R)))/2 = .75(9.8/R) = 9.8(.75)
Average = 75%
Work = Force * Distance = ((7.35 *440) * 6400)Joules
(sorry, no calc)